Is it possible to convert an SARIMA model to a continuous model?

If so, what is the methodology to do that?

  • $\begingroup$ The bridge that you're looking for is called the discretization scheme. The first order scheme is named after Euler with a second order named after Milstein. You may find some difficulty as in some sense you are going "backwards" as most modeling is done in SDEs first then simulated with Euler or Milstein discretizations. Also, there has been much written about calibrating the parameters of different kinds of SDEs that often make use of these schemes. $\endgroup$ – user25064 Jun 3 '16 at 12:06
  • $\begingroup$ I have not come across a continuous version of SARIMA before but that does not mean that it cannot be created. The answer by @Richard may be generalized with a bit of calculus. It is a bit like writing a proof where you have to work backward and think "What is the SDE such that when I apply the Euler discritization it looks like an AR/MA/ARMA process?" $\endgroup$ – user25064 Jun 3 '16 at 12:35
  • $\begingroup$ Excellent input, thank you. You gave one direction of the bridge (Cont. -> Discrete). Can you give me the inverse? $\endgroup$ – Dionysios Georgiadis Jun 3 '16 at 12:37
  • $\begingroup$ To certain degree the question of discretization is more natural than of "making it continuous". Of cours "finding the continuous counerpart" is a valid aim. $\endgroup$ – Ric Jun 3 '16 at 12:43
  • $\begingroup$ It actually appears that Euler discretization is not enough because it only includes one time step, same with Milstein. There is no SDE that exists that when you apply an Euler discretization you will get an AR(n) process. $\endgroup$ – user25064 Jun 3 '16 at 13:00

Let's look at the formula for an ARMA(p, q) model

$$ X_t = c + \sigma z_t + \sum_{i=1}^p \varphi_i X_{t-i} + \sigma\sum_{i=1}^{q}\theta_i z_{t-i} $$

where $z_t \in \mathcal{N}(0,1)$ for all $t$. Transform this into the continuous time counterpart below.

$$ X_t = c + \sigma W_t + \int_0^{T_p}\varphi(v)X_{t-v} dv + \sigma\int_0^{T_q}\theta(v)W_{t-v}dv $$

Where $T_{p,q}$ are the width of the lookback window (in years) for the AR and MA terms. The sequence of coefficients have been replaced by continuous functions of time $\varphi, \theta : \mathbb{R_+} \mapsto \mathbb{R}$. Interpolation methods can be used to go from the discrete $\varphi_i$ to the continuous $\varphi(v)$.

A problem that I see with this is that I believe the $z_t$ should probably not be replaced with $W_t$ but something more like "$\sqrt{dt}dW_t$ " but I am not sure how to formalize that. I am also not sure about how to write out the total derivative for this SDE $dX_t$ any help with that would be appreciated.

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You find the connection between an AR(1) and an Ornstein-Uhlenbeck process here in QSE if you search.

Taken the description form here (which is just the way to do it):

$$ X_{n+1} = c + a X_n + b \varepsilon_k $$ and by setting $c=\theta \mu \Delta t, a=−\theta \Delta t$ and $b =\sigma \sqrt{\Delta t}$ you will get the discrete time approximation $$ X_{n+1} = \theta(\mu - X_n)\Delta t + \sigma \varepsilon_k\sqrt{\Delta t}, $$ which corresponds to the continuous time OU-process.

If you add a seasonal process then you have SAR(1).

Thus AR(1) + seasonal component in discrete time is $$ dX_t = a (b-\mu) dt + dB_t + dS_t $$ where $S_t$ is a periodic function. This would be the solution for the AR(1) case and no seasonal AR or MA component.

If we think of the dependence structure in higher order AR processes I wonder whether there is a continuous time model for AR(n) for $n>1$.

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  • $\begingroup$ I am aware of the AR1 OU connection, but I am interested in a more general mapping. Thanks for the very clear explanation though! $\endgroup$ – Dionysios Georgiadis Jun 3 '16 at 10:13
  • $\begingroup$ Do you know how to find the continuous time equivalent of AR(n)? $\endgroup$ – Ric Jun 3 '16 at 11:29
  • $\begingroup$ I do not, I would consider that part of the answer to the initial question. $\endgroup$ – Dionysios Georgiadis Jun 3 '16 at 12:32
  • $\begingroup$ Maybe this is the central part. AR(1) is clear. Normal seasonality is clear. What about AR(n)? This could be a stand-alone question-. $\endgroup$ – Ric Jun 3 '16 at 12:35
  • $\begingroup$ I see your point. $\endgroup$ – Dionysios Georgiadis Jun 3 '16 at 12:41

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